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Leokris [45]
3 years ago
9

Calculate the mass of water producd from the reaction of 126g of pentaboron noahydride with 192g of moleculAR OXYGEN

Chemistry
1 answer:
Sindrei [870]3 years ago
5 0

Answer:

81.08g of H_{2}O will be produced.

Explanation:

Write down the balanced chemical equation:

B_5H_9 + O_2 ⇒ B_2O_3+H_2O

2B_5H_9+12O_2 ⇒ 5B_2O_3+9H_2O

Determine the limiting reagent:

B_5H_9 :-    126/63.12646 = 1.995993 mol

               1.995993/2 = 0.9979965

O_2 :-         192/31.9988 = 6.000225 mol

               6.000225/12 = 0.50001875

Therefore, O_2, is the limiting reagent.

Use stoichiometry ratios to determine the number of moles of water produced:

         O_2         :        H_2O

         12          :          9

  6.000225   :       4,500168756328362

Use the mole formula to calculate the mass of water produced:

n=\frac{m}{M} \\m=nM\\m=(4.500168756328362)(18.01528)\\m=81.08g

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